Can a type alias be created for more than one parameter of a class or function with multiple type parameters?

When using Vue, there are situations where a generic function may require 3, 4, or even 5 type parameters. Is it possible to create a type alias for these parameters in order to avoid typing them out repeatedly? Something like this perhaps:

// Example of a generic function with multiple type parameters
function verboseGenericFunc<T, U, V extends T>(thing: T): V {
  // Implementation using T, U, V
  return thing as V;
};

// Type alias representing the tuple of types required by the function
type VerboseTypeParams = [ConcreteT, ConcreteU, ConcreteV];

// Using the type alias to specify all required type parameters
const v = verboseGenericFunc<...VerboseTypeParams>(thing);

Answer №1

To create a type object that acts as an alias for the type arguments provided to a templated function, you can use a conditional type setup in TypeScript. This type object will resolve to 'undefined' if incorrect generic types are supplied when defining the type alias.

type VerboseTypeParams<T,U,V> = V extends T? {"T":T, "U":U, "V":V}: undefined;

type concrete = VerboseTypeParams<string,number, string>; // Result corresponds to the desired type

Subsequently, you can extract individual types from this type alias object within a wrapper function.

// Wrapper function with generics
function verboseGenericFunc(thing: concrete["T"] ):concrete["V"] {
  return verboseGenericFuncVue<concrete["T"],concrete["U"],concrete["V"]>(thing);
};

The purpose of this wrapper function is to call the actual target function while abstracting away the need to provide any type arguments explicitly. TypeScript will flag errors if incompatible wrappers are created or incorrect types are passed to the corresponding wrapper functions.

// Function with multiple type parameters (e.g., from Vue)
function verboseGenericFuncVue<T, U, V extends T>(thing: T): V {
  // Perform operations involving T, U, V
  return thing as V;
};

type concreteBad = VerboseTypeParams<string,number, number>; // Resolves to 'undefined'

// Incorrect Generic function wrapper triggering TypeScript error due to key lookup issues with 'concreteBad' being undefined
function verboseGenericFuncBad(thing: concreteBad["T"] ):concreteBad["V"] {
  return verboseGenericFuncVue<concreteBad["T"],concreteBad["U"],concreteBad["V"]>(thing);
};

const v1 = verboseGenericFunc("1"); // Successful call to wrapped function without specifying types
const v2 = verboseGenericFunc(1); // Type error raised by TypeScript since a string was expected according to 'concrete'

Playground Link

Answer №2

Similar concept to the response from MacD but with a slight variation in implementation: utilize the helper type as the type parameter for your functions.

type InfoType<T = any, U = any, V extends T = any> = { T: T, U: U, V: V };

// Function that is generic with multiple type parameters
function customGenericFunction<TI extends InfoType>(item: TI['T']): TI['V'] {
  // Perform operations involving T, U, V
  return item as TI['V'];
};

// Creating an alias for the tuple of types that adhere to the required parameter interfaces
type CustomTypeParameters = InfoType<number, string, 5>;

// Using the alias to fulfill all the type parameters
const x: 5 = customGenericFunction<CustomTypeParameters>(12020);

// One-time type declaration.
const y: 'hi' = customGenericFunction<InfoType<string, null, 'hi'>>('hello');

Playground Link

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